# 代数数

${\displaystyle \mathbb {N} \subseteq \mathbb {Z} \subseteq \mathbb {Q} \subseteq \mathbb {R} \subseteq \mathbb {C} }$

## 定义

${\displaystyle z}$复数。如果存在正整数${\displaystyle n}$，以及${\displaystyle n+1}$个有理数${\displaystyle q_{0},q_{1},\cdots ,q_{n}}$，并且${\displaystyle q_{n}\neq 0}$，使得：
${\displaystyle q_{n}z^{n}+\cdots +q_{1}z+q_{0}=0}$

## 性质

${\displaystyle \mathbb {Z} _{n}^{k}[X]=\left\{a_{0}+a_{1}X+\cdots +a_{n}X^{n};\;\;a_{0},a_{1},\cdots ,a_{n}\in \mathbb {Z} ,\;a_{n}\neq 0,\;|a_{0}|+|a_{1}|+\cdots +|a_{n}|=k\right\}}$

${\displaystyle \mathbb {Z} _{n}^{k}[X]}$中多项式的任何系数至多有${\displaystyle 2k+1}$个可能性，最高次项系数至多有${\displaystyle 2k}$个可能性，因此这样的多项式个数不超过${\displaystyle 2k(2k+1)^{n}}$。每个多项式至多有${\displaystyle n}$个根。如果将所有${\displaystyle \mathbb {Z} _{n}^{k}[X]}$中多项式的根的集合记为${\displaystyle {\mathcal {A}}_{n}^{k}}$，则${\displaystyle {\mathcal {A}}_{n}^{k}}$的元素个数不超过${\displaystyle 2nk(2k+1)^{n}}$，即为有限集。

${\displaystyle \mathbb {Z} [X]=\mathbb {Z} \bigcup _{n\in \mathbb {Z} ^{+},k\in \mathbb {Z} ^{+}}\mathbb {Z} _{n}^{k}[X].}$

${\displaystyle {\mathcal {A}}=\bigcup _{n\in \mathbb {Z} ^{+},k\in \mathbb {Z} ^{+}}{\mathcal {A}}_{n}^{k}.}$

${\displaystyle \mathbb {Z} ^{+}\times \mathbb {Z} ^{+}}$是可数集。集合${\displaystyle {\mathcal {A}}}$是可数个有限集的并集，因此是可数的。

## 参考文献

• Artin, Michael, Algebra, Prentice Hall, 1991, ISBN 0-13-004763-5, MR1129886
• Ireland, Kenneth; Vosen, Michael, A Classical Introduction to Modern Number Theory, Graduate Texts in Mathematics 84 Second, Berlin, New York: Springer-Verlag, 1990, ISBN 0-387-97329-X, MR1070716
• G. H. Hardy and E. M. Wright 1978, 2000 (with general index) An Introduction to the Theory of Numbers: 5th Edition, Clarendon Press, Oxford UK, ISBN 0-19-853171-0
• Lang, Serge, Algebra, Graduate Texts in Mathematics 211 4th, Springer-Verlag, 2004, ISBN 0-387-95385-X
• Orestein Ore 1948, 1988, Number Theory and Its History, Dover Publications, Inc. New York, ISBN 0-486-65620-9 (pbk.)